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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1/9g/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 9G / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

A Hamming code decoder succeeds on a seven-bit block exactly when there are zero or one errors. Its radius-one decoding regions are disjoint, so an error of weight at least two cannot decode back to the original codeword. Thus

q = (1 - p)^{7} + 7 p (1 - p)^{6} = 1 - 21 p^{2} + O (p^{3}), Q = q^{N} ≃ (1 - 21 p^{2})^{N} ≃ e^{- 21 N p^{2}} .

(1)

Numerically

Q \approx 0.97929

, so only about

1.0212

transmissions are expected. Even accounting for seven rather than four bits per letter, expected bits sent are

7 N / Q \approx 7148

, compared with

4 N / P \approx 218830

. The Hamming method is substantially better under the stated error detection and independent retransmission assumptions.

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